Testing Change of Basis in Linear Operator

[psholtz]

I just want to test/verify my knowledge of change of basis in a linear operator.. (it's not a homework question).

Suppose I have linear operator mapping R^2 into R^2, and expressed in the canonical basis (1,0), (0,1). Suppose (for the sake of discussion) that the linear operator is given by:

$$A=\left(\begin{array}{ccc}5 & 6 \\ 7 & 8\end{array}\right)$$

Suppose now that I want to express this linear operator in a different basis, say (2,1), (1,2). To do this, I first apply the linear operator to the two (original) basis vectors:

$$A \cdot \left(\begin{array}{c}1 \\ 0 \end{array}\right) = \left(\begin{array}{c}5 \\ 7\end{array}\right)$$

$$A \cdot \left(\begin{array}{c}0 \\ 1 \end{array}\right) = \left(\begin{array}{c}6 \\ 8\end{array}\right)$$

To find the linear operator A' in the new basis, we express these two vectors in terms of the new basis vectors. Solving the equation for x1, x2 gives:

$$\left(\begin{array}{cc}2 & 1 \\ 1 & 2 \end{array}\right) \left(\begin{array}{c}x_1 \\ x_2\end{array}\right) = \left(\begin{array}{c}5 \\ 7\end{array}\right)$$

$$x_1 = 1, x_2=3$$

and similarly:

$$\left(\begin{array}{cc}2 & 1 \\ 1 & 2 \end{array}\right) \left(\begin{array}{c}x_1 \\ x_2\end{array}\right) = \left(\begin{array}{c}6 \\ 8\end{array}\right)$$

$$x_1 = 4/3, x_2=10/3$$

So the expression for the linear operator in the new basis will be given by:

$$A' = \left(\begin{array}{cc}1 & 4/3 \\ 3 & 10/3\end{array}\right)$$

Is this correct?

 

[Outlined]

Let
$$B = \left(\begin{array}{cc}2 & 1 \\ 1 & 2 \end{array}\right)$$

Now

$$A' = B^{-1}AB = \frac{1}{3}\left(\begin{array}{cc}10 & 11 \\ 28 & 29 \end{array}\right)$$